參數(shù)資料
型號(hào): 73M1903-IGV
廠商: TERIDIAN SEMICONDUCTOR CORP
元件分類: 消費(fèi)家電
英文描述: SPECIALTY CONSUMER CIRCUIT, PQFP32
封裝: TQFP-32
文件頁數(shù): 38/45頁
文件大小: 449K
代理商: 73M1903-IGV
73M1903
Modem Analog Front End
DATA SHEET
Page: 43 of 45
2005 TERIDIAN Semiconductor Corporation
Rev 1.4
18
1
7
1
Dnco2
Nnco2
Dnco1
Nnco1
)
1
18
(
)
7
1
(
7
18
/
=
=
Fxtal
Fvco
, where Nnco1 and Nnco2 must be < or equal to 8.
The ratio, Nnco1/Dnco1 = 1/7, is used to form a divide ratio for the NCO in prescaler and Nnco2/Dnco2 =
1/18 for the NCO in the PLL.
Prescaler NCO: From Nnco1/Dnco1 = 1/7,
Pdvsr = Integer [ Dnco1/Nnco1 ] = 7;
Prst[2:0] = Nnco1 – 1 = 0; this means NO fractional divide. It always does ÷7. Thus Pseq becomes
“don’t care” and is ignored.
Pseq = {x,x,x,x,x,x,x,x} = xxh.
PLL NCO: From Nnco2/Dnco2 = 1/18,
Ndvsr = Integer [ Dnco2/Nnco2 ] = 18;
Nrst[2:0] = Nnco2 – 1 = 0; this means NO fractional divide. It always does ÷18. Thus Pseq becomes
“don’t care” and is ignored.
Nseq = {x,x,x,x,x,x,x,x} = xxh.
Example 2:
Crystal Frequency = 24.576MHz; Desired Sampling Rate, Fs = 10.971kHz=2.4kHz x 8/7 x4
Step 1. First compute the required VCO frequency, Fvco, corresponding to
Fs = 2.4kHz x 8/7 x 4 =10.971kHz.
Fvco = 2 x 2304 x Fs = 2 x 2304 x 2.4kHz x 8/7 x 4 = 50.55634MHz.
Step 2. Express the required VCO frequency divided by the Crystal Frequency as a ratio of two integers.
This is initially given by :
MHz
Fxtal
Fvco
576
.
24
4
8/7
2.4kHz
2304
2
/
=
.
After a few rounds of simplification this ratio reduces to:
18
1
35
4
Dnco2
Nnco2
Dnco1
Nnco1
)
1
18
(
)
35
4
(
/
=
=
Fxtal
Fvco
, where Nnco1 and Nnco2 must be < or equal to 8.
The ratio, Nnco1/Dnco1 = 4/35, is used to form a divide ratio for the NCO in pre-scaler and Nnco2/Dnco2
=1/18 for the NCO in the PLL.
相關(guān)PDF資料
PDF描述
73M1903-IM SPECIALTY CONSUMER CIRCUIT, QCC32
73M1903-IGV/F SPECIALTY CONSUMER CIRCUIT, PQFP32
73M1903-IGV SPECIALTY CONSUMER CIRCUIT, PQFP32
73M1903-IM SPECIALTY CONSUMER CIRCUIT, QCC32
73M1903-IM/F SPECIALTY CONSUMER CIRCUIT, PQCC32
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